# Adiabatic Expansion of an Ideal Gas

####
**Adiabatic Expansion of an Ideal Gas**

**❒**A process carried in a vessel whose walls are perfectly insulated so that no heat can pass through them, is said to be adiabatic.

**❒**In adiabatic process there is no heat exchange between a system and surroundings, and q = 0.

**❒**According to the First law:

ΔE =
q – w = 0 – w

ΔE =
– w ...

**(1)****❒**Since the work is done at the expense of internal energy, the internal energy decreases and the temperature falls.

**❒**Consider 1 mole of an ideal gas at pressure P and a volume V. For an infinitesimal increase in volume dV at pressure P, the work done by the gas is –PdV. The internal energy decreases by dE.

**❒**According to equation (1):

dE =
– PdV ...

**(2)**
By definition of molar heat capacity at constant volume:

dE =
C

_{v}dT ...**(3)**
From (2) and (3):

C

_{v}dT = – PdV
For an ideal gas:

P =
RT/V

and hence:

Integrating between T

_{1}, T_{2}and V_{1}, V_{2}and considering C_{v}to be constant:
Since R = C

_{p}– C_{v }, this equation may be written as:
The ratio of C

_{p}to C_{v}is often written as γ :
and equation (4) thus becomes:

_{2}/V

_{1}to V

_{1}/V

_{2}and taking antilogarithms:

We can also eliminate the temperature by making use of the ideal
gas relationship:

####
**Comparison between Isothermal and Adiabatic Expansions**

Boyle’s law describes pressure-volume relations of an ideal gas
under isothermal conditions (T, constant). This is similar to the relation
derived for adiabatic expansion.

PV =
constant (Boyle’s law)

PV

^{γ}= constant (Adiabatic expansion)
γ for an ideal monoatomic gas = 1.67.

The difference between the two processes is : in an isothermal
process, temperature of a system remains constant while in an adiabatic
process, temperature must change.

###
**Explanation:**

In an isothermal process heat is absorbed to make up for the work
done by the gas in expansion and the temperature remains unchanged.

On the other hand, adiabatic expansion takes place at the expense
of internal energy which decreases and the temperature falls. For the same reason,
the curve for the adiabatic process is steeper than that for the isothermal process.

####
**Work done In adiabatic Reversible Expansion**

**Step (1): Value of VdP from adiabatic equation**

For an adiabatic process:

PV

^{γ}= constant
Differentiating it, we have:

γ PV

^{γ – 1}dV + V^{γ}dp = 0
Dividing by V γ – 1, we get:

γ
PdV + VdP = 0

VdP
= – γ PdV ...

**(1)****Step (2): Value of VdP from ideal gas equation**

For 1 mole of an ideal gas:

PV =
RT

Complete differentiation gives:

PdV
+ VdP = RdT

VdP
= RdT – PdV ...

**(2)****Step (3): Substitution**

Substituting the value of VdP from (1) in (2) we get:

RdT
– PdV = – γ PdV

RdT
= P (1 – γ) dV

**Step (4): Integration**

Integrating from T

_{1}, V_{1}to T_{2}, V_{2}with γ constant:
When T

_{2}> T_{1}, w_{max}is negative because 1 – γ is negative. This means that work is done on the
gas.

On the other hand, when T

_{2}< T_{1}, w_{max}is positive which means that work is done by the gas.####
**Solved Problem**

**Problem (1):Calculate w for the adiabatic reversible expansion of 2 moles of an ideal gas**

**at 273.2 K and 20 atm to a final pressure of 2 atm.**

**Solution:**

Given:

C

_{p}= 5R/2, mole^{–1}deg^{–1}
C

_{v}= 3R/2, mole^{–1}deg^{–1}
R = 8.314J
mole

^{–1}deg^{–1}**Step 1:**To calculate the value of T

_{2}, the final temperature, using the equation:

(T

_{2}/ T_{1})^{γ}= (P_{2}/ P_{1})^{γ –1}
Substituting the value of γ in (1):

(T

_{2}/ 273.2)^{5/3}= (2 / 20)
Solving it, we get:

T

_{2}= 108.8 K**Step 2:**To calculate maximum work under adiabatic conditions:

The work done under adiabatic conditions may be obtained by
calculating decrease in internal

Energy:

w =
– ΔE = – nC

_{v}(T_{2}– T_{1})
= –
2 × 3 / 2 × 8.314 (108.8 – 273.2)

=
4100 J = 4.1 kJ

**Problem (2): At 25ºC for the combustion of 1 mole of liquid benzene the heat of reaction at constant pressure is given by:**

**What would be the heat of reaction at constant volume?**

**Solution:**

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